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If $r \ge 3$ is an integer, show that $4x^r = 3y^2 + 1$ does not have positive integer solutions $(x, y)$ except for $(1, 1)$. (I am not sure whether this is an open problem; in any case, it is a slightly stronger but probably more familiar form of this question.)

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Eric Neyman
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    perhaps working in $$Z(\frac{1+\sqrt{-3}}{2})$$ would work – B. S. Jun 26 '16 at 07:25
  • Thanks, but could you clarify what technique(s) I might want to use, or give a link to a resource that would explain how working over $Z \left( \frac{1 + \sqrt{-3}}{2} \right)$ would help? – Eric Neyman Jun 26 '16 at 07:33
  • any book on algebraic number theory should work, provided you know some abstract algebra. – B. S. Jun 26 '16 at 07:35
  • a nice, introductory read is T.Nagell "Introduction to the theory of numbers", or Hardy and Wright's book http://matematica.cubaeduca.cu/medias/pdf/842.pdf – B. S. Jun 26 '16 at 07:37
  • start at page 178 of hardy and wright's book, which explains it – B. S. Jun 26 '16 at 07:47

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